# Cardinality of the set of continuous functions

Suppose $$(X,\tau)$$ and $$(Y,\sigma)$$ are topological spaces. Let $$F(X,Y)$$ be the set of continuous functions $$X\rightarrow Y$$. I want to compute the cardinality of $$F(X,Y)$$. It depends not only on cardinalities $$|X|$$ and $$|Y|$$ but on the topologies as well: just imagine what happens if $$X$$ or $$Y$$ is discrete.

Question: is it possible to write some "topological invariants" $$|\tau|$$ and $$|\sigma|$$ that yield the following explicit formula: $$|F(X,Y)| = \Theta (|X|,|Y|,|\tau|,|\sigma|)$$

I am sorry for being sloppy: I really need to clarify what a formula is. Ideally, $$|\tau|$$ would be a collection of cardinals and the formula will be given in the cardinal arithmetic. However, I would be glad to have any other method of computing the cardinality. Thus, "the formula" may be something more general.

• There is a continuum that only allow constant mappings and identity to be continuous. May be relevant – erz Feb 8 '20 at 20:58
• If you allow proper class invariants, then the answer is yes, because you can essentially encode the whole of F into the invariant in that case. – Joel David Hamkins Feb 8 '20 at 21:25
• You can also trivially do it if you have global choice, from which you get a definable well-ordering of $V$ of order type $Ord$. For any topological space you find the lowest index $\alpha$ of a homeomorphic topological space, and then your 'cardinal invariant' is $\aleph_\alpha$. – James Hanson Feb 8 '20 at 21:32
• @BugsBunny The point is that a lot of information can be coded in cardinals and under certain common set theoretic assumptions you can actually code the homeomorphism type of $(X,\tau)$ by a unique cardinal. So if you're too permissive about what $|\tau|$ and $\Theta$ are, the answer to your question can become very sensitive to set theoretic assumptions and even trivial under certain specific assumptions. – James Hanson Feb 8 '20 at 23:47
• Next: in <a href="eudml.org/doc/170472"> Wann sind alle stetige abbildungen in $Y$ konstant? </a> Herrlich shows that for every $T_1$-spae $Y$ there is a regular space $X$ such that every continuous map from $X$ to $Y$ is constant, so $|F(X,Y)|=|Y|$. There is no bound on the cardinality of $X$. In particular of $Y$ is the space of rationals $F(X,Y)$ is countable for very large $X$. – KP Hart Feb 10 '20 at 12:36